由网友(哭到世界为我痛)分享简介：什么是最快的算法（链接到C或C ++的例子是凉）检查一个小方阵（小于16×16元）？是单数（不可逆，DET = 0）What is the fastest algorithm (a link to C or C++ example would be cool) to check if a small square m...

什么是最快的算法（链接到C或C ++的例子是凉）检查一个小方阵（小于16×16元）？是单数（不可逆，DET = 0）

What is the fastest algorithm (a link to C or C++ example would be cool) to check if a small square matrix (<16*16 elements) is singular (non-invertible, det = 0) ?

推荐答案

最好的办法是计算条件数通过SVD并检查是否大于1 /ε，其中，小量的机器precision。

Best way is to compute the condition number via SVD and check if it is greater than 1 / epsilon, where epsilon is the machine precision.

如果您允许假阴性（即一个矩阵是有缺陷的，但是你的算法可能无法检测它），你可以使用MAX（a_ii）/分钟（a_ii）公式从维基百科的文章作为条件数的代理，但你必须先计算QR分解（该公式适用于三角矩阵）：A = QR其中R正交，然后COND（A）= COND（Q）。也有技术来计算的Q为O（N）的操作的条件数，但也存在更复杂的

If you allow false negatives (ie. a matrix is defective, but your algorithm may not detect it), you can use the max(a_ii) / min(a_ii) formula from the Wikipedia article as a proxy for the condition number, but you have to compute the QR decomposition first (the formula applies to triangular matrices): A = QR with R orthogonal, then cond(A) = cond(Q). There are also techniques to compute the condition number of Q with O(N) operations, but there are more complex.